Understanding the "unroling" step in the proof of the policy gradient theorem
The unrolling step is due to the fact you end up with an equation that you can keep expanding indefinitely.
Note that we start with calculating $\nabla v_\pi(s)$ and arrive at
$$\nabla v_\pi(s) = \sum_a\left[ \nabla \pi(a|s) q_\pi(s,a) + \pi(a|s) \sum_{s'}p(s'|s,a) \nabla v_\pi (s') \right]\;,$$
which contains a term for $\nabla v_\pi(s')$. This is a recursive relationship, similar to the bellman equation, so we can substitute in a term for $\nabla v_\pi(s')$ which will be a term similar just with $\nabla v_\pi(s'')$. As I mentioned, we can do this indefinitely which leads us to
$$\nabla v_\pi(s) = \sum_{x \in \mathcal{S}} \sum_{k=0}^\infty \mathbb{P}(s\rightarrow x, k, \pi) \sum_a \nabla \pi(a|x) q_\pi(x,a)\;.$$
We need the term $\sum_{x \in \mathcal{S}} \sum_{k=0}^\infty \mathbb{P}(s\rightarrow x, k, \pi)$ because we want to take an average over the state space, however due to unrolling there are many different $s_t$'s that we need to average over (this comes from the $s',s'',s''',...$ in the unrolling) so we also need to add the probability state of transitioning from state $s$ to $x$ in $k$ time steps, where we sum over an infinite horizon due to the repeated unrolling.
If you are wondering what happens to the terms $\pi(a|s)$ and $p(s'|s,a)$ terms and why they are not explicitly shown in this final form, it is because this is exactly what the $\mathbb{P}(s\rightarrow x, k, \pi)$ represents. The average over all possible states accounts for the $p(s'|s,a)$ and the fact that we follow policy $\pi$ in the probability statement accounts for the $\pi(a|s)$.